数列{an}中,a1=1,an,an+1是方程x2-(2n+1)x+1bn=0的两个根,则数列{bn}的前n项和Sn等于()A.n2n+1B.nn+1C.12n+1D.1n+1-数学

题目简介

数列{an}中,a1=1,an,an+1是方程x2-(2n+1)x+1bn=0的两个根,则数列{bn}的前n项和Sn等于()A.n2n+1B.nn+1C.12n+1D.1n+1-数学

题目详情

数列{an}中,a1=1,an,an+1是方程x2-(2n+1)x+
1
bn
=0
的两个根,则数列{bn}的前n项和Sn等于(  )
A.
n
2n+1
B.
n
n+1
C.
1
2n+1
D.
1
n+1
题型:单选题难度:中档来源:不详

答案

由题意可得an+an+1=2n+1
∴an=n
anan+1=class="stub"1
bn
bn=class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

Sn=b1+b2+…+bn
=(1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…class="stub"1
n
-class="stub"1
n+1
)

=1-class="stub"1
n+1
=class="stub"n
n+1

故选B.

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